It’s possible to have an equiangular quadrilateral, i.e. whose sides are geodesics (the analogue of “straight line” on a sphere). The Gauss-Bonnet theorem implies their total interior angle is greater than 2pi, so four right angles can’t work.
Notice that each side is a segment of a great circle, i.e. a circle that divides the sphere in half. That’s what it means for a path to be a geodesic on the sphere.
Good point. Four equal angles, then, although they will each have to be greater than 90 degrees.
I don’t think that would work for just 4 lines? I think you have to have arcs, not straight lines
It’s possible to have an equiangular quadrilateral, i.e. whose sides are geodesics (the analogue of “straight line” on a sphere). The Gauss-Bonnet theorem implies their total interior angle is greater than 2pi, so four right angles can’t work.
Here’s an interactive demo of quadrilaterals on the sphere: https://geogebra.org/m/q83rUj8r
Notice that each side is a segment of a great circle, i.e. a circle that divides the sphere in half. That’s what it means for a path to be a geodesic on the sphere.